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Dr Roger Bennett R.A.Bennett@Reading.ac.uk Rm. 23 Xtn. 8559

Thermal Physics PH2001. Dr Roger Bennett R.A.Bennett@Reading.ac.uk Rm. 23 Xtn. 8559. Lecture 4. Kinetic Theory of Gases. The Maxwell-Boltzmann Velocity Distribution We assume an ideal gas (non interacting massive point particles undergoing elastic collisions).

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Dr Roger Bennett R.A.Bennett@Reading.ac.uk Rm. 23 Xtn. 8559

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  1. Thermal Physics PH2001 Dr Roger Bennett R.A.Bennett@Reading.ac.uk Rm. 23 Xtn. 8559 Lecture 4

  2. Kinetic Theory of Gases • The Maxwell-Boltzmann Velocity Distribution • We assume an ideal gas (non interacting massive point particles undergoing elastic collisions). • Imagine a box of gas with atoms bouncing around inside. Each one with a velocity (u=vx,v=vy,w=vz) in Cartesian coordinates. • The probability that a molecule has its x component of velocity in the range du about u is defined as • Pu = f(u)duwhere f(u) is (the currently unknown) probability density function. • Similarly Pv = f(v)dv and Pw = f(w)dw

  3. v u w Kinetic Theory of Gases • The probability that u will be in a range du andv will be in a range dv andw will be in a range dw is • Puvw = Pu Pv Pw = f(u)f(v)f(w) du dv dw • Now the probability distribution (f(u)f(v)f(w))must be spherically symmetric as we have no preferred direction:-

  4. Kinetic Theory of Gases • The spherical symmetry of the surface means that if we move around on a contour of constant probability we don’t change the probability: d(f(u)f(v)f(w))= 0 • Using the product rule again f’(u)du f(v)f(w) + f’(v)dv f(u)f(w) + f’(w)dw f(u)f(v) = 0 Equation 4.1 • But we have a spherical surface so u2 + v2 + w2 = constant udu + vdv + wdw = 0 Equation 4.2 • For arbitrary du and dv we can rearrange to: dw = (-udu – vdv) / w Equation 4.3

  5. Kinetic Theory of Gases • Substuting dw in eqn. 4.1 with dw from eqn. 4.3 and rearranging gives: {f’(u)/f(u) – (u/w) f’(w)/f(w))}du + {f’(v)/f(v) – (v/w) f’(w)/f(w))}dv = 0 • As du and dv are arbitrary the terms in brackets must be zero: • f’(u)/f(u) – (u/w) f’(w)/f(w) = 0 • f’(v)/f(v) – (v/w) f’(w)/f(w) = 0 • So • f’(u)/(uf(u)) = f’(w)/(wf(w)) = f’(v)/(vf(v)) = -B • Bis an unknown constant • f’(u) = - Buf(u) (and similarly for v and w)

  6. Kinetic Theory of Gases • f’(u) = - Buf(u) (and similarly for v and w) • f(u) = Ae-1/2Bu2 This gives the shape of the probability distribution for the velocity in each direction • f(v) = Ae-1/2Bv2 and f(w) = Ae-1/2Bw2 • A is an arbitrary scaling constant

  7. Kinetic Theory of Gases • f(u) = Ae-1/2Bu2 • Find A by normalising

  8. Kinetic Theory of Gases • A more useful quantity is the speed distribution. • Puvw=f(u)f(v)f(w)dudvdw is the probability of finding an atom in du at uand in dv at vand in dw at w.

  9. Kinetic Theory of Gases

  10. Kinetic Theory of Gases • This is the Maxwell-Boltzman Distribution for the total speed c in a gas. • We still need to understand B.

  11. Kinetic Theory of Gases • To find B we need to relate our microscopic understanding to the macroscopic. • We know PV = (2/3) U = nkT = (2/3) n<mc2/2> • Average kinetic energy per molecule = 3/2kT • What is the average speed and kinetic energy of the Maxwell-Boltzmann distribution? • (3/2)kT = (1/2)m<c2> = (3/2) m/B • B = m/kT

  12. Maxwell-Boltzman Distribution for the speed c.

  13. Thermal Physics PH2001 Dr Roger Bennett R.A.Bennett@Reading.ac.uk Rm. 23 Xtn. 8559 Lecture 5

  14. Equations of State • We have found that PV = nkT = NRT = (2/3) U • When three of the five state variables P,V, N (or n), T or U are specified then the remaining two are determined and the state of the system is known. • If the variables do not change with time it is an equilibrium state. • We are most often interested in changes of state as a result of external action such as compressing a gas, stretching a rubber band, cooling etc.Such actions are processes.

  15. Equations of State • For the ideal Gas: PV = nkT = NRT = (2/3) U • We can sketch this equation of state as a surface:

  16. V A dV Changes of State • We have already looked at one change of state – adiabatic compression. On a P-V diagram this can be visualised as:- dw = -PdV • The blue dots define initial and final states with unique values of P and V and hence T (or U).

  17. Changes of State • If we do this carefully and slowly through a succession of intermediate equilibrium states we can show that the work done on the system is area under curve.

  18. Changes of State • The succession of intermediate equilibrium states has a special name – a quasistatic process. • It is important as it means we always lie on the equation of state during the process. • If we moved from initial to final states rapidly pressure and temperature gradients would occur, and the system would not be uniquely described by the equation of state.

  19. V A dV Changes of State • If we don’t do this carefully and slowly what happens? • Consider a compression where the wall moves quite fast, with velocity uB. • A molecule with:- • Recoils with u  u+2uB

  20. Changes of State • Thus the molecule has gained excess energy: • If uB<<u then we recover our standard result ke=uB(2mu)= uB(momenta to wall per impact) U(total/area/sec)= uB (total momenta/area/sec) U(total/area A/time dt) = uBPAdt = -PdV = dW • If uB<<u is not true then 2muB2additionally increases the internal energy U as uB2is always positive irrespective of the sign of uB. • This additional term is physically manifested as the flow of heat.

  21. Changes of State • Thus in general we write the conservation of energy as: dU = đW + đQ This is the First Law of Thermodynamics in differential form • Note the subtle difference between the d in dU and đ in đW. dU is a perfect differentialbecause U is a state function and uniquely determined. đW and đQ areimperfect differentialsbecause W and Q are path functions that depend on the path taken.

  22. Changes of State – Thermodynamic Processes • Adiabatic a process with no heat transfer into or out of the system. Therefore, the system may have work done on it or do work itself. • Isochoric a process undertaken at constant volume. If the volume is constant then the system can do no work on its surroundings đW = 0. • Isobaric a process undertaken at constant pressure. Q, U and W can all vary but finding W is easy as W = -P(V2-V1).

  23. Changes of State – Thermodynamic Processes • Isothermal processes are undertaken at constant temperature. This is achieved by coupling the system to a reservoir or heat bath. Heat may flow in or out of the system at will but the temperature is fixed by the bath. In general isothermal processes U, Q and W can all vary. For the special cases, such as the ideal gas, where U only depends on the temperature the heat entering the system must equal the work done by the system.

  24. Changes of State – Thermodynamic Processes • These processes for a fixed quantity of ideal gas can be shown on a single P-V indicator diagram.

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